Welcome 2016! Part 32

$a_n$ can be written as $n(n+1)(n+2)...(n+9)$ , i.e. it is the product of the 10 consecutive integers starting from $n$ . Let $k$ be the smallest positive integer for which the rightmost non-zero digit of $a_k$ is odd. We can write $a_k$ as:

$a_k = 10^pb$

where $2 \nmid b$ , i.e $p$ is the largest power of 2 that divides $a_k$ . Secondly, $5^p\mid a_k$ . Since we want to find the smallest positive integer for which this is true we need to find the smallest value for $p$ for which this is true.

To calculate $a_n$ , we are multiplying 10 consecutive integers. Hence, we will always have 5 of them even, at least two of them divisible by 4 and at least one divisible by 8. That means for all $n$ , $2^8 \mid a_n$ . To find $k$ , we want $5^8 \mid a_k$ and $2^9 \nmid a_k$ .

We always have two multiples of 5 in 10 consecutive integers, one odd and one even. (Note: only one of them can be divisible 25.) As long as one of them is divisible by $5^7$ , we will have $5^8 \mid a_k$ . Since we are looking for the smallest value, we start setting the odd multiple to $5^7 = 78125$ . This means $k \ge 78116$ . If we choose $k = 78116$ , we have 3 multiples of 4 in the next 10 integers, namely 78116, 78120, 78124 and hence $2^9 \mid a_{78116}$ . So we choose $k=78117$ . Therefore:

$a_k = 78117\times78118\times78119\times78120\times78121\times78122\times78123\times78124\times78125\times78126$

Simplifying we get, $a_k = 78117\times39059\times78119\times1953\times78121\times39061\times78123\times19531\times39063\times5^8\times2^8$

Clearly we have $2^9\nmid a_k$ and thus the last non-zero digit is odd. In order find the last non-zero digit, $d$ , we just multiply the last digits to get:

$d = 7\times9\times9\times3\times1\times1\times3\times1\times3 \mod 10 = 9$

which is our final answer.